A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there is a 53% chance that the player will miss both shots, a 31% chance that he will make one of the shots, and a 16% chance that he will make both shots.
a. Construct the appropriate probability distribution. (Round your answers to 2 decimal places.)
b. What is the probability that he makes no more than one of the shots? (Round your answer to 2 decimal places.)
c. What is the probability that he makes at least one of the shots? (Round your answer to 2 decimal places.)
a. To construct the probability distribution, we assign probabilities to the three possible outcomes: missing both shots (53%), making one shot (31%), and making both shots (16%). We add up these probabilities to ensure they sum to 1.
Probability of missing both shots: 53% = 0.53
Probability of making one shot: 31% = 0.31
Probability of making both shots: 16% = 0.16
Therefore, the probability distribution is:
P(X = 0) = 0.53
P(X = 1) = 0.31
P(X = 2) = 0.16
b. To find the probability that he makes no more than one of the shots, we need to sum the probabilities of making 0 shots and making 1 shot.
P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.53 + 0.31 = 0.84
Therefore, the probability that he makes no more than one of the shots is 0.84.
c. To find the probability that he makes at least one of the shots, we need to subtract the probability of making 0 shots from 1.
P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.53 = 0.47
Therefore, the probability that he makes at least one of the shots is 0.47.
a. To construct the probability distribution, we can assign the corresponding probabilities to each possible outcome.
Let's define:
X = the number of shots the player makes
P(X = 0) = 0.53 (misses both shots)
P(X = 1) = 0.31 (makes one shot)
P(X = 2) = 0.16 (makes both shots)
The probability distribution is as follows:
X | P(X)
---------
0 | 0.53
1 | 0.31
2 | 0.16
b. To find the probability that he makes no more than one of the shots, we need to calculate:
P(X ≤ 1) = P(X = 0) + P(X = 1)
P(X ≤ 1) = 0.53 + 0.31 = 0.84
Therefore, the probability that he makes no more than one of the shots is 0.84.
c. To find the probability that he makes at least one of the shots, we need to calculate:
P(X ≥ 1) = 1 - P(X = 0)
P(X ≥ 1) = 1 - 0.53 = 0.47
Therefore, the probability that he makes at least one of the shots is 0.47.
To construct the probability distribution for a basketball player's free throws, we need to consider the given probabilities for various outcomes:
Let's represent:
X = Number of shots made
a. The appropriate probability distribution for X is:
X | P(X)
---------------
0 | 0.53 (53% chance of missing both shots)
1 | 0.31 (31% chance of making one shot)
2 | 0.16 (16% chance of making both shots)
This probability distribution summarizes the chances of each possible outcome for the number of shots made.
b. To find the probability that the player makes no more than one shot, we need to calculate the sum of the probabilities for X = 0 and X = 1:
P(X ≤ 1) = P(X = 0) + P(X = 1)
= 0.53 + 0.31
= 0.84
Therefore, the probability that the player makes no more than one shot is 0.84, or 84%.
c. To find the probability that the player makes at least one shot, we need to calculate the complement of the probability that the player makes no shots (X = 0):
P(X ≥ 1) = 1 - P(X = 0)
= 1 - 0.53
= 0.47
Therefore, the probability that the player makes at least one shot is 0.47, or 47%.